Difference between revisions of "1984 AIME Problems/Problem 8"

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This reduces <math>\theta</math> to either 120 or 160. But <math>\theta</math> can't be 120 because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\theta=160</math>.
 
This reduces <math>\theta</math> to either 120 or 160. But <math>\theta</math> can't be 120 because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\theta=160</math>.
 
== See also ==
 
== See also ==
* [[1984 AIME Problems/Problem 7 | Previous problem]]
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{{AIME box|year=1984|num-b=7|num-a=9}}
* [[1984 AIME Problems/Problem 9 | Next problem]]
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* [[AIME Problems and Solutions]]
* [[1984 AIME Problems]]
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* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]

Revision as of 13:24, 6 May 2007

Problem

The equation $\displaystyle z^6+z^3+1$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in thet complex plane. Determine the degree measure of $\theta$.

Solution

If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value 1 and argument of the form $40m^\circ$ for integer $m$.


This reduces $\theta$ to either 120 or 160. But $\theta$ can't be 120 because if $r=\cos 120^\circ +i\sin 120^\circ$, then $r^3=1$ and $r^6+r^3+1=3$, a contradiction. This leaves $\theta=160$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AIME Problems and Solutions
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